A proton, a neutron, an electron and an $\alpha$ particle have same energy. If $\lambda$p, $\lambda$n, $\lambda$e and $\lambda$a are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following :
Solution
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{(2 m K E)}}$
<br/><br/>$\lambda \propto \frac{1}{\sqrt{m}}$ as all the particles have same KE.
<br/><br/>Since $m_e < m_p < m_{\mathrm{n}} < m_\alpha$
<br/><br/>$\lambda_e > \lambda_p > \lambda_{\mathrm{n}} > \lambda_\alpha$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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